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QUESTION


Consider the following modification of the classic Buffon's needle
experiment: two needles of the same length, $\ell$, fused at right angles at
their centres, are thrown on a horizontal ruled floor, with parallel
lines at a distance $d=\ell$ apart. Let $X$ and $Y$ be binary variables, indicating
whether each of the needles crosses a line. Let  $Z = X + Y$ be the
number of times that this cross arrangement intercepts a line.

\begin{description}
\item[(i)] Find Prob(Z = 0), Prob(Z = 1) and Prob(Z = 2).

\item[(ii)] Find the variance of $Z$.

\item[(iii)]Using the fact that var$(Z) = $var$(X) +
$var$(Y) + 2$cov$(X,Y)$, find cov$(X,Y)$.

\item[(iv)] Comment on the use of this cross arrangement
as a variance reduction technique.
\end{description}

\bigskip

ANSWER

Simulation

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\put(4,2.5){\line(-2,-3){.2}} \put(4.5,2.2){$\theta$}
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\begin{description}
\item[(i)]

$P(Z=2)=(A+B)/(\pi l/2)$

$P(Z=0)=(C+D)/(\pi l/2)$

$P(Z=1)=1-P(Z=0)-P(Z=2)$

\begin{eqnarray*}
A+B&=&2\times\left\{\int_0^{\frac{\pi}{4}}\frac{l}{2}\sin\theta\,d\theta
+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{l}{2}
\sin\left(\frac{\pi}{2}-\theta\right)\,d\theta\right\}\\
&=&l\times\left[\frac{}{}-\cos\theta\right]_0^{\frac{\pi}{4}}
+l\times\left[\cos\left({\frac{\pi}{2}}-\theta\right)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}\\
&=&l\times\left[-\frac{\sqrt2}{2}+1+1-\frac{\sqrt2}{2}\right]=l\left(2-\sqrt2\right)
\end{eqnarray*}

$P(Z=2)=\left[l(2-\sqrt2)\right]/(\pi l/2)=2(2-\sqrt2)/\pi$

\bigskip
\begin{eqnarray*}
C+D&=&2\times\left\{\frac{\pi l}{4}
-\int_0^{\frac{\pi}{4}}\frac{l}{2}\sin\left(\frac{\pi}{2}-\theta\right)\,d\theta
-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{l}{2}
\sin\theta\,d\theta\right\}\\
&=&l\times\left[\frac{\pi}{2}-\cos\left(\frac{\pi}{2}-\theta\right)\right]_0^{\frac{\pi}{4}}
+l\times\left[\frac{}{}\cos\theta\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}\\
&=&l\times\left(\frac{\pi}{2}-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\right)
=l\times\left(\frac{\pi}{2}-\sqrt2\right)
\end{eqnarray*}

$\ds P(Z=0)=\left[l\left(\frac{\pi}{2}-\sqrt2\right)\right]/(\pi
l/2)=\left(\pi-2\sqrt2\right)/\pi$

$\ds
P(Z=1)=1-\frac{2(2-\sqrt2)}{\pi}-\frac{(\pi-2\sqrt2)}{\pi}=\frac{4(\sqrt2-1)}{\pi}$

Hence $P(Z=0)=0.0997,\quad P(Z=1)=0.527,\quad P(Z=2)=0.373$

\item[(ii)]

$E(Z)=1.273,\quad E(Z^2)=2.019,\quad {\rm var}(Z)=0.398$

\item[(iii)]

$X$ and $Y$ have the same distribution:

\begin{eqnarray*}
P(X=1)&=&\frac{2}{\pi
l/2}\int_0^{\frac{\pi}{2}}\frac{l}{2}\sin\theta\,
d\theta\\&=&\frac{2}{\pi}\left[\frac{}{}-\cos\theta\right]_0^{\frac{\pi}{2}}
=\frac{2}{\pi}\\
P(X=0)&=&1-\frac{2}{\pi}
\end{eqnarray*}

$\ds E(X)=\frac{2}{\pi},\quad {\rm
var}(X)=\frac{2}{\pi}\left(1-\frac{2}{\pi}\right)={\rm var}(Y)$

Therefore ${\rm cov}(X,Y)=(0.398-2\times0.231)/2=-0.032$

\item[(iv)]

Observe that $\ds
{\rm var}\left(\frac{Z}{2}\right)=\frac{1}{4}{\rm var}(Z)=0.0995$

If we throw one needle twice on the floor,

$${\rm var}\left(\frac{X_1+X_2}{2}\right)=\frac{{\rm
var}(X)}{2}=0.1155$$
The size of the cross arrangement therefore not only speeds up the
process but also reduces the variance


\end{description}


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